Find the equation of the tangent plane to the surface $z=x^2+2y^3$ at the point $(1,1,3)$. I think that it is $z=2x+6y-5$. Is that right?
$\endgroup$ 51 Answer
$\begingroup$The normal to the surface $x^2+2y^3-z=0$ is parallel the gradient: $(2x,6y^2,-1)$
At $(1,1,3)$, this is $(2,6,-1)$. The tangent plane is perpendicular to this vector. That is, $$ (2,6,-1)\cdot(x,y,z)=2x+6y-z=5\tag{1} $$ where $5$ is gotten by plugging $(1,1,3)$ into $2x+6y-z$. $(1)$ is the equation of the tangent plane.
$\endgroup$
